# Back propagation algorithm of Neural Network : XOR training

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Ashikur on 22 Jan 2012
Commented: Greg Heath on 9 Nov 2018
c=0;
wih = .1*ones(nh,ni+1);
who = .1*ones(no,nh+1);
while(c<3000)
c=c+1;
for i = 1:length(x(1,:))
for j = 1:nh
netj(j) = wih(j,1:end-1)*double(x(:,i))+wih(j,end)*1;
outj(j) = 1./(1+exp(-1*netj(j)));
end
% hidden to output layer
for k = 1:no
netk(k) = who(k,1:end-1)*outj'+who(k,end)*1;
outk(k) = 1./(1+exp(-1*netk(k)));
delk(k) = outk(k)*(1-outk(k))*(t(k,i)-outk(k));
end
% back proagation for j = 1:nh s=0; for k = 1:no s = s+who(k,j)*delk(k); end
delj(j) = outj(j)*(1-outj(j))*s;
s=0;
end
for k = 1:no
for l = 1:nh
who(k,l)=who(k,l)+.5*delk(k)*outj(l);
end
who(k,l+1)=who(k,l+1)+1*delk(k)*1;
end
for j = 1:nh
for ii = 1:ni
wih(j,ii)=wih(j,ii)+.5*delj(j)*double(x(ii,i));
end
wih(j,ii+1)=wih(j,ii+1)+1*delj(j)*1;
end
end
end
// The code above, I have written it to implement back propagation neural network, x is input , t is desired output, ni , nh, no number of input, hidden and output layer neuron. I am testing this for different functions like AND, OR, it works fine for these. But XOR is not working.
// Training x = [0 0 1 1; 0 1 0 1] // Training t = [0 1 1 0]
// who -> weight matrix from hidden to output layer
// wih -> weight matrix from input to hidden layer
// Can you help ?

Greg Heath on 24 Jan 2012
It would help immensely if you posted a code that could be cut, pasted, run, and yield a numerical answer. Then we could begin by matching our results with yours.
It would also help immensely if you replaced the loops with vectorized code.
Greg
Greg Heath on 24 Jan 2012
If you initialized weights randomly, you could see if it is
an initialization problem.
Have you noticed the loop accidentally included in the backpropagation comment?
Greg

Greg Heath on 25 Jan 2012
close all, clear all, clc
x = [0 0 1 1; 0 1 0 1]
t = [0 1 1 0]
[ni N] = size(x)
[no N] = size(t)
nh = 2
% wih = .1*ones(nh,ni+1);
% who = .1*ones(no,nh+1);
wih = 0.01*randn(nh,ni+1);
who = 0.01*randn(no,nh+1);
c = 0;
while(c < 3000)
c = c+1;
% %for i = 1:length(x(1,:))
for i = 1:N
for j = 1:nh
netj(j) = wih(j,1:end-1)*x(:,i)+wih(j,end);
% %outj(j) = 1./(1+exp(-netj(j)));
outj(j) = tansig(netj(j));
end
% hidden to output layer
for k = 1:no
netk(k) = who(k,1:end-1)*outj' + who(k,end);
outk(k) = 1./(1+exp(-netk(k)));
delk(k) = outk(k)*(1-outk(k))*(t(k,i)-outk(k));
end
% back propagation
for j = 1:nh
s=0;
for k = 1:no
s = s + who(k,j)*delk(k);
end
delj(j) = outj(j)*(1-outj(j))*s;
% %s=0;
end
for k = 1:no
for l = 1:nh
who(k,l) = who(k,l)+.5*delk(k)*outj(l);
end
who(k,l+1) = who(k,l+1)+1*delk(k)*1;
end
for j = 1:nh
for ii = 1:ni
wih(j,ii) = wih(j,ii)+.5*delj(j)*x(ii,i);
end
wih(j,ii+1) = wih(j,ii+1)+1*delj(j)*1;
end
end
end
h = tansig(wih*[x;ones(1,N)])
y = logsig(who*[h;ones(1,N)])
e = t-round(y)
Hope this helps.
Greg

#### 1 Comment

Ashikur on 25 Jan 2012
Hi Greg, Thank you so much for the effort and time you have put on this. I am really grateful to you.
But the problems still remains, your modification (basically random initialization of weights and using tan sigmoid for hidden layer) sometimes works sometimes not. And I must know your motivation for not using logsig everywhere. Are there any explanations?
I want to share the whole code which is now in better shape. And you are right about the vectorization, I will do that once the code is working.
Here is two functions one for training and another for simulation.
You can try to run like this, and see sometimes it works sometimes not, I have added your suggestion in code.
>> net = biasBackProp(2,3,1,1,0.0001,[0 0 1 1;0 1 0 1],[0 1 1 0])
>> bpSim([0;1],net)
>> bpSim([1;1],net)
>> bpSim([1;0],net)
>> bpSim([0;0],net)
The functions can be copied from here,

Greg Heath on 27 Jan 2012
It is well known that successful deterministic training depends on a lucky choice of initial weights. The most common approach is to use a loop and create Ntrial (e.g., 10 or more) nets from different random initial weights. Then choose the best net.
It is also well known that an odd bounded monotonically increasing activation function like TANSIG is the choice of preference for hidden layers because it does not restrict the polarity of the layer variables. It works even better when the input is shifted to have zero mean.
You can check the superiority of TANSIG and zero-mean yourself. You can also search the comp.ai.neural-nets FAQ and archives to find both agreement and numerical experiments.
For most real world problems the best choice for number of hidden nodes, H, is not known apriori. That is why I have posted many examples using a double loop: An outer loop over H and an inner loop over Ntrials random weight initializations. For examples, search the newsgroup using the keywords
heath clear Ntrials
Hope this helps.
Greg

Ashikur on 27 Jan 2012
Thanks, you are awesome :)
Imran Babar on 6 May 2013
Thank you very much a nice example of MLP BP NN and very easy to understand. Though I am a novice in this field but I am now clear in the programming idea

Imran Babar on 8 May 2013
Dear sir I want to use the same code for the following data set
Input dataset=[1 1 1 2;1 1 2 2;1 2 2 2; 2 2 2 2] Output=[5 6 7 8]
but it is always generating output as given below
1 1 1 1
I tried my best but unable to understand how may I get these results

#### 1 Comment

Greg Heath on 10 May 2013
Your outputs are not within the range of logsig.
Either normalize your outputs to fit in {0,1)
or
change your output activation function (e.g., 'purelin')

Sohel Ahammed on 4 Jul 2015
Ok. If i Want to test it, how i have to change. Ex: input : 1 0 expected output : 1 (From learing).

dsmalenb on 17 Oct 2018
Am I missing something here but I don't see any bias neurons. Maybe this is why you are getting some inputs to work and others not?

Greg Heath on 17 Oct 2018
The ones are multiplied by the bias weights which are automatically learned with the others.
dsmalenb on 17 Oct 2018
I'm sorry but that statement does not make much sense to me. Biases are added to shift the values within the activation function. Multiplying by 1 does nothing.
Greg Heath on 9 Nov 2018
The 1 is a placeholder which is multiplied by a learned weight.
Hmm, I've been using that notation for decades and this is the 1st question re that that I can remember.
Greg